第6章代数表达式和身份–练习6.6 |套装1
问题11.如果x – y = 7且xy = 9，则求出x 2 + y 2的值
解决方案：
Given in the question x – y = 7 and x y = 9
By squaring on both sides
(x – y)2 = 72
x2 + y2 – 2xy = 49
x2 + y2 – 2 (9) = 49 (since x y=9)
x2 + y2 – 18 = 49
x2 + y2 = 49 + 18
x2 + y2 =67
问题12。如果3x + 5y = 11并且xy = 2，则找到9x 2 + 25y 2的值
解决方案：
Given in the question 3x + 5y = 11 and x y = 2
on squaring on both sides
(3x + 5y)2 = 112
(3x)2 + (5y)2 + 2(3x)(5y) = 121
9x2 + 25y2 + 2 (15xy) = 121 (given x y=2)
9x2 + 25y2 + 2(15(2)) = 121
9x2 + 25y2 + 60 = 121
9x2 + 25y2 = 121-60
9x2 + 25y2 = 61
问题13.查找以下表达式的值：
(i)当x = 7/4时为16x 2 + 24x + 9
We will using the formula (a + b)2 = a2 + b2 + 2ab
=(4x)2 + 2 (4x) (3) + 32
=(4x + 3)2
Putting when x = 7/4
=[4 (7/4) + 3]2
=(7 + 3)2
=100
(ii)当x = 11和y = 4/3时为64x 2 + 81y 2 + 144xy
We will using the formula (a + b)2 = a2 + b2 + 2ab
=(8x)2 + 2 (8x) (9y) + (9y)2 (8x + 9y)
Putting when x = 11 and y = 4/3
=[8 (11) + 9 (4/3)]2
=(88 + 12)2
=(100)2
=10000
(iii)当x = 2/3和y =¾时为81x 2 + 16y 2 – 72xy
We will using the formula (a + b)2 = a2 + b2 + 2ab
=(9x)2 + (4y)2 – 2 (9x) (4y)
=(9x – 4y)2
Putting x = 2/3 and y = 3/4
=[9 (2/3) – 4 (3/4)]2
=(6 – 3)2
=32
=9
问题14.如果x + 1 / x = 9，则找到x4 + 1 / x4的值。
解决方案：
Given in the question x + 1/x = 9
squaring both sides
(x + 1/x)2 = (9)2
x2 + 2 × x × 1/x + (1/x)2 = 81
x2 + 2 + 1/x2 = 81
x2 + 1/x2 = 81 – 2
x2 + 1/x2 = 79
Now again when we square on both sides
(x2 + 1/x2)2 = (79)2
x4 + 2 × x2 × 1/x2 + (1/x2)2 = 6241
x4 + 2 + 1/x4 = 6241
x4 + 1/x4 = 6241- 2
x4 + 1/x4 = 6239
问题15.如果x + 1 / x = 12，则求出x – 1 / x的值。
解决方案：
Given in the question x + 1/x = 12
When squaring both sides
(x + 1/x)2 = (12)2
x2 + 2 × x × 1/x + (1/x)2 = 144
x2 + 2 + 1/x2 = 144
x2 + 1/x2 = 144 – 2
x2 + 1/x2 = 142
When subtracting 2 from both sides
x2 + 1/x2 – 2 × x × 1/x = 142 – 2
(x – 1/x)2 = 140
x – 1/x = √140
问题16。如果2x + 3y = 14且2x – 3y = 2，则求出xy的值。 [提示：使用(2x + 3y) 2 –(2x-3y) 2 = 24xy]
解决方案：
2x + 3y = 14 equation (1)
2x – 3y = 2 equation (2)
Now square both the equations and subtract equation (2) from equation (1)
(2x + 3y)2 – (2x – 3y)2 = (14)2 – (2)2
4×2 + 9y2 + 12xy – 4×2 – 9y2 + 12xy = 196 – 4
24 xy = 192
xy = 8
问题17。如果x 2 + y 2 = 29且xy = 2，则求出值
(i)x + y
(ii)x – y
(iii)x 4 + y 4
解决方案：
(i) x + y
x2 + y2 = 29 (Given in the question)
x2 + y2 + 2xy – 2xy = 29
(x + y) 2 – 2 (2) = 29
(x + y) 2 = 29 + 4
x + y = ± √33
(ii) x – y
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x – y)2 + 2 (2) = 29
(x – y)2 + 4 = 29
(x – y)2 = 25
(x – y) = ± 5
(iii) x4 + y4
x2 + y2 = 29
Squaring both sides
(x2 + y2)2 = (29)2
x4 + y4 + 2x2y2 = 841
x4 + y4 + 2 (2)2 = 841
x4 + y4 = 841 – 8
= 833
问题18.必须将以下每个表达式加什么才能使其成为一个完整的正方形?
(i)4x 2 – 12x + 7
(2x)2 – 2 (2x) (3) + 32 – 32 + 7
(2x – 3)2 – 9 + 7
(2x – 3)2 – 2
(ii)4x 2 – 20x + 20
(2x)2 – 2 (2x) (5) + 52 – 52 + 20
(2x – 5)2 – 25 + 20
(2x – 5)2 – 5
问题19：简化：
i)(x – y)(x + y)(x 2 + y 2 )(x 4 + y 4 )
(x2 – y2) (x2 + y2) (x4 + y4)
[(x2)2 – (y2)2] (x4 + y4)
(x4 – y4) (x4 – y4)
[(x4)2 – (y4)2]
x8 – y8
(ii)(2x – 1)(2x +1)(4x 2 +1)(16x 4 +1)
[(2x)2 – (1)2] (4x2 + 1) (16x4 + 1)
(4x2 – 1) (4x2 + 1) (16x4 + 1) 1
[(4x2)2 – (1)2] (16x4 + 1) 1
(16x4 – 1) (16x4 + 1) 1
[(16x4)2 – (1)2] 1
256x8 – 1
(iii)(7m – 8n) 2 +(7m + 8n) 2
(7m)2 + (8n)2 – 2(7m)(8n) + (7m)2 + (8n)2 + 2(7m)(8n)
(7m)2 + (8n)2 – 112mn + (7m)2 + (8n)2 + 112mn
49m2 + 64n2 + 49m2 + 64n2
grouping the similar expression
98m2 + 64n2 + 64n2
98m2 + 128n2
(iv)(2.5p – 1.5q) 2 –(1.5p – 2.5q) 2
on expansion
(2.5p)2 + (1.5q)2 – 2 (2.5p) (1.5q) – (1.5p)2 – (2.5q)2 + 2 (1.5p) (2.5q)
6.25p2 + 2.25q2 – 2.25p2 – 6.25q2
grouping the similar expression
4p2 – 6.25q2 + 2.25q2
4p2 – 4q2
4 (p2 – q2)
(v)(m 2 – n 2 m) 2 + 2m3n 2
On expansion using (a + b)2 formula
(m2)2 – 2 (m2) (n2) (m) + (n2m)2 + 2m3n2
m4 – 2m3n2 + (n2m)2 + 2m3n2
m4+ n4m2 – 2m3n2 + 2m3n2
m4+ m2n4
问题20：表明：
(i)(3x + 7) 2 – 84x =(3x – 7) 2
LHS => (3x + 7)2 – 84x
We will using the formula (a + b)2 = a2 + b2 + 2ab
(3x)2 + (7)2 + 2 (3x) (7) – 84x
(3x)2 + (7)2 + 42x – 84x
(3x)2 + (7)2 – 42x
(3x)2 + (7)2 – 2 (3x) (7)
(3x – 7)2 = R.H.S
(ii)(9a – 5b) 2 + 180ab =(9a + 5b) 2
LHS => (9a – 5b)2 + 180ab
We will using the formula (a + b)2 = a2 + b2 + 2ab
(9a)2 + (5b)2 – 2 (9a) (5b) + 180ab
(9a)2 + (5b)2 – 90ab + 180ab
(9a)2 + (5b)2 + 9ab
(9a)2 + (5b)2 + 2 (9a) (5b)
(9a + 5b)2 = R.H.S
(ⅲ)(4M / 3 – 3N / 4)2 +的2mn = 16立方米/ 9 + 9N 2/16
LHS => (4m/3 – 3n/4)2 + 2mn
(4m/3)2 + (3n/4)2 – 2mn + 2mn
(4m/3)2 + (3n/4)2
16/9m2 + 9/16n2 = R.H.S
(iv)(4pq + 3q) 2 –(4pq – 3q) 2 = 48pq 2
LHS => (4pq + 3q)2 – (4pq – 3q)2
(4pq)2 + (3q)2 + 2 (4pq) (3q) – (4pq)2 – (3q)2 + 2(4pq)(3q)
24pq2 + 24pq2
48pq2 = RHS
(v)(a – b)(a + b)+(b – c)(b + c)+(c – a)(c + a)= 0
LHS =>(a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)
By using the identity (a – b) (a + b) = a2 – b2
(a2 – b2) + (b2 – c2) + (c2 – a2)
a2 – b2 + b2 – c2 + c2 – a2
0 = R.H.S